`(2^100+3^100+4^100+5^100)/7`<br>
`=(2*2^99+(9^(1/2))^100+2^200+5*5^99)/7`<br>
`=(2*(2^3)^33+9^50+2^2*(2^3)^66+5*(5^3)^33)/7`<br>
`=(2*8^33+9^50+2^2*8^66+5*(125)^33)/7`<br>
`=(2*(1)^33+(2)^50+2^2*(1)^66+5*(-1)^33)/7`<br>
`=(2*(1)^33+2^2*(8)^16+2^2*(1)^66+5*(-1)^33)/7`<br>
`=(2*(1)^33+2^2*(1)^16+2^2*(1)^66+5*(-1)^33)/7`<br>
`=(2+4+4-5)/7 = 5/7`<br>
So, the remainder will be `5` when the given number is divided by `7`.<br>